What is the integral of (2x)/(k^2) ?

The integral of (2x)/(k^2) is (x^2)/(k^2)+C

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integral of (2x)/(k^2)

\int \frac{2 x}{k ^{2}} d x

\frac{x ^{2}}{k ^{2}} + C

Solution steps

\int \frac{2 x}{k ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{k ^{2}} \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= \frac{2}{k ^{2}} \cdot \frac{x ^{2}}{2}

Simplify

\frac{2}{k ^{2}} \cdot \frac{x ^{2}}{2} : \frac{x ^{2}}{k ^{2}}

= \frac{x ^{2}}{k ^{2}}

Add a constant to the solution

= \frac{x ^{2}}{k ^{2}} + C

x \to \infty lim (x^{2} - 4 x)

f (x) = 4 sin^{3} (x)

y^{^{'}} + \frac{2}{t} y = \frac{sin ( t )}{t}

\int (\frac{1}{1 - x ^{2}}) d x

\int \frac{x - 3}{x ^{2} - 49} d x

What is the integral of (2x)/(k^2) ?
The integral of (2x)/(k^2) is (x^2)/(k^2)+C